# Does Gauss's law not hold when there are charges outside of the Gaussian surface?

For example, if I have a gaussian surface and a negative charge somewhere outside that gaussian surface, there's going to be a non-zero electric flux.

However, According to gauss's law, electric flux = $\frac{Q}{E}$ and $Q$ is $0$ because there are no charges inside the surface?

• The flux is zero. It's positive in some places and negative in others. – knzhou Jan 17 '16 at 3:44
• What about the electric field? That's not 0 – StackOverflowOfficial Jan 17 '16 at 3:57

The electric field is not zero. Flux =$\int \mathbf{E.ds}$ is zero not the electric field itself. for a negative single charge electric filed is $\mathbf{E}=\frac{Q}{4\pi \epsilon_0 r^2}(-\hat{r})$. The flux is zero because $\mathbf{ds}$ has different direction with respect to $\hat{r}$ at different point on the surface and the total flux is therefore zero.